# Thread: Help with integrating

1. ## Help with integrating

the problem$\displaystyle \int 2t/(t-3)^2$
solving with partial fractions$\displaystyle \frac{a}{t-3} + \frac{a}{(t-3)^2}$
$\displaystyle t = 3, b=6$
$\displaystyle t=0, a =2$
$\displaystyle \int \frac{2}{t-3} + \int \frac{6}{(t-3)^2}$
$\displaystyle = 2lnt-3 + \int \frac{6}{(t-3)^2}$
but I don't know how to solve for $\displaystyle \int \frac{6}{(t-3)^2}$

2. $\displaystyle \displaystyle \int{\frac{6}{(t - 3)^2}\,dt} = \int{6(t - 3)^{-2}\,dt}$.

Solve using a $\displaystyle \displaystyle u$ substitution.

3. Oh wow! I knew it was something simple, completely forgot that I can change the exponent into -2 = /
I didn't do a U subsitution though, all I did was take out the 6 and integrate with (t-3) as one number, which I guess is like substituting for u.
Thanks!
I got -6/t-3

4. Originally Posted by dorkymichelle
I didn't do a U subsitution though, all I did was take out the 6 and integrate with (t-3) as one number, which I guess is like substituting for u.
You did, in fact, get the right answer. But you should be familiar with the steps of this method when things are not so easy.
$\displaystyle \displaystyle u = t - 3 \implies du = dt$

So your integral turns into:
$\displaystyle \displaystyle 6 \int (t - 3)^{-2} dt = 6 \int u^{-2} du = 6(-1)u^{-1} + C = -6(t - 3)^{-1} + C$

-Dan

$\displaystyle \displaystyle -6/t - 3 = -\frac{6}{t} - 3$